The core insight is sorting by start time and using a min-heap to track when each worker becomes free next.
Model the problem as interval graph coloring where jobs are vertices and edges connect overlapping jobs; the minimum number of workers equals the maximum clique size (maximum overlap). Use a sweep-line algorithm to process events in time order, assigning jobs to available workers via a min-heap of end times, and track assignments. Return the worker ID for each job.
Pro tip: When explaining, emphasize that the greedy assignment is optimal because at any time, the number of overlapping jobs is a lower bound, and the algorithm achieves it. Also, mention that if jobs are already sorted by start time, the sweep is O(n log n) due to heap operations.
Confirm input format (list of jobs with start/end times, possibly IDs) and output (worker assignment per job). Discuss assumptions: times are integers, jobs are half-open intervals [start, end), and workers are interchangeable.
Select a sweep-line approach: sort events (start and end) by time, with end events processed before start events at the same time to allow reuse. Use a min-heap to track end times of busy workers and a pool of available worker IDs.
Iterate through events: on a start event, if heap not empty and top end <= current start, free that worker; assign the job to an available worker (or new if none) and push its end time to heap. On an end event, mark worker as free (or handle via heap).
State time complexity: O(n log n) due to sorting and heap operations. Space complexity: O(n) for heap and assignment array. Mention that the number of workers used equals the maximum overlap.
Walk through a small example (e.g., jobs: [1,3], [2,4], [3,5]) to show assignment and verify no overlaps. Discuss edge cases: empty list, all jobs overlapping, jobs with same start/end times.
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